You have 12 coins, one weighs slightly less or slightly more than the others. Using an equal arm balance and only making three weighings determine which one is different and whether it is slightly less or slightly more. Warning: the fact that you don’t know whether it is less or more is a major problem in the
solution.

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you divide the coins into 3 groups of 4 take two of the groups and measure them then if they’re equal then the coin must be in the third group if one weighs less the coin must be there. then now knowing which group it is separate the pennies into groups of two one should weigh less take that group and measure the last two pennies and the one that weights less is the penny you were looking for

No problem — feel free to email me solutions at sjm1@williams.edu. We’re still trying to figure out exactly what we want to do on the webpage. For now, it’s more general discussion about the riddles (such as the moat should be rectangular for one of them), or how these might be used in a class.

First measure 1~6. 123 in one side and 456 in the other.
If You see a difference, for example if the (123) is heavier and (456) is lighter,
U know there is a heavier one or lighter one in one of both sides.

Therefore you take (157) and put it in one side and (428) in one side.
Now you have 3 different numbers from different groups.
Numbers 1 and 2 comes from the group that weighed heavier
Numbers 5 and 4 comes from the group that weighed less
Finally numbers 7 and 8 comes from the group that has the normal coins.

Now if 157 still weights more than 428, U know that 1 is heavier or 4 is lighter.
Therefore now you take 1 or 4 and compare it with 7~12 (any coin is alright).
Now you know the answer.

If 157 weights less than 428, u know that either 5 or 2 is the answer.
Therefore you compare it with 7~12 (any is fine).

If 123 and 456 weights the same
You now measure 789 101112.
If 789 weighs more than 101112
You put 7,10,11 in one side and 8,1,2 in the other side.
( You notice we left out 9 and 12).
If (71011) still weights more and (812) weights less,
then we know that 7 weights more.
If (71011) weights less and (812) weights more,
Then we know that 10 or 11 weights less or 8 weights more.
Therefore you just measure 10 and 11. if they’re equal u know that 8 weights more.
If they’re not, well u know that one of them weights lesser.

1st Step:
– Take 6 coins at random from the 12
– Split them into two groups of three and weigh across the balance.
If they are in balance, set aside these 6 coins as the “good” pile (and the other 6 as the “bad” pile). If they are not in balance, then set the random 6 as the “bad” pile.

2nd step:
– Take 3 coins at random from the “bad” pile and 3 from the “good” pile and weigh them across the scale.
– If the “good” and “bad” piles are not in balance, then the randomly selected 3 from the “bad” has a defective coin. This will also tell you whether the defective coin is less or more. If in balance, then the 3 not selected at random contain the defective coin.

3rd step:
– From the defective 3, take 2 coins at random and weigh across the balance.
– The defective coin will weigh more or less (as judged by step 2). If in balance, then the 3rd non-selected coin is the defective.

However, I’ve now realized that unless you’re able to conclusively prove whether the defective coin weighs less or more in step 2 (by getting an unbalanced scale), you won’t be able to decide conclusively on the defective coin in step 3.

actually you’ll get this problem if you got a balaned scale (not unbalanced scale) .. what if you remember when testing 3 on 3 from the bad pile, which one weigh less (or more) … now if you choose the same set of 3 that you have tested in step one you will be able to decide if the coin weighs less or more .. I’am right ?

Step one: Weigh two groups of 5 coins and leave 2 coins off to the side:
Step two: Take 4 coins from the lighter/heavier side from from Step One and place 1 coint off the side, weighing only 2 groups of two.
Step three: Finally, weigh only 1 coin on each side using the two lighter/heavier coints from Step two.

If when doing Step 1 the weighs would have been equal, you could have known it was between the two you originally left to the side.

If when doing step two the weights were equal, you could have weighed the side pair from step one to find it, or if those were equal simply inferred that the single coin left to the side AFTER step one was the special coin.

After doing step 1 and having the weights be off, you would have known that neither of the two originally set-asside coins were NOT the special coin.

My possible solution is to divide the 12 by 2 and place 6 coins on each side, then the heaviest redivide by 2 until you place 1 coin on each side of the balance and holding one separate. If the one is a different weight it would be indicated but if it balances out the one that is not placed on the balance is the odd weighted one. That is when I got stuck…

– If 1 is not equal to 2 then we weigh 1 and 3 — Weigh No.2
If they are equal then the coin is in group 2 and we know if 2 is lighter or heavier than other groups
From group 2 we take two coins out of it and weigh them — Weigh No. 3 — if they equal that the coin which has been left out is the coin we are looking for if not than one of the two weighed coins will be the different one (we know already if it is lighter or heavier)

If 1 and 3 are different the the coin is in group 1 and we know if it heavier or lighter depending on the result of the previous weigh and we do the same exercise for the three coins remaining

– If 1 = 2 then we weigh 3 and 1 — Weigh No. 2
If 1 is not equal to 3 then we know that 3 contains the coin and we know if it is lighter or heavier
We take group 3 and we take two coins out of it and weigh them Weigh No. 3 if they equal that the coin which has been left out is the coin we are looking for if not than one of the two weighed coins will be the different one (if 3 is lighter than 1 then we take the lighter coin and vice versa)

The problem is if 1 is equal to 2 and is equal to 3 then we cannot know if the coin in the group 4 is lighter or heavier.
Can you please share with me the solution by email?
Thx

This is what i have been thinking so far…. am I on the right lines?:
Start by halving the coins into two 6s. Choose one of these sixes and divide into half and place 3 coins on one side, and 3 coins on the other. If the coins balance equally, then these coins can be disguarded, or if they dont, then keep them to re-weigh???

in the first attempt we have 6 on each side. the one heavier would have the heavy coin. Now from the heavier batch of coins we select four coins and weigh with 2 on each side. if they are equal then on the last attempt we check the other 2 coins the one heavier would be our coin. If they aren’t then we would way, then we would take the set of two coins which is heavier and weigh with each coin in each side of the arm balance. the heavier coin would ours.

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I don’t know if I just suck at those coin-weighing riddles or if this one is above ‘medium’, but it was a really tough nut to crack. I did it, eventually, but not until I’ve spent almost two hours getting my brain evaporated through my ears. This really is a challenging one since the strategy is different depending on the result of the first weighing.

Weigh 3 coins against 3 coins. If they balance, they are all good coins, if they don’t, the ones not being weighed are all good coins. Now there are 6 good coins and a group of 6 containing one bad coin. Weigh 3 good coins against 3 potential bad coins. If they balance, the remaining 3 contain the bad coin. If they don’t balance, the bad coin is in that group of 3 and it will be obvious if it’s lighter or heavier. Now there are 9 good coins. From the remainig 3 coins, take 2 and weigh them against each other. If they balance, the third one is the bad one. If they don’t, the bad coin is the one that is lighter or heavier (depending on the results of the second weighing).

put 4 on one side of scale and remember weight. put four on other side. there the same so discard those 8. put 2 on same side of scale and see if that weight is half of the first 4.
if so discard those 2. put 1 on one side of scale, if it’s weight is half of 2 youve found your coin

There is an interesting twist to this problem. The weighings have to be independent, i.e., you have to specify upfront what is going to be weighed in each weighing and then, solely based on the results of the weighings, determine which penny is different from the others and whether it is lighter or heavier than the rest.

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